Steel roof trusses residential10/23/2023 N Ed/N C,Rd = 12.896/156.475 = 0.0824 < 1 Therefore section is ok for uniform compression. Resistance of the member to uniform compression Thus, the section satisfies both of the conditions. Radius of gyration (axis y-y) r i = 1.5 cmĬonsidering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.72 cm 2Įquivalent tension area for welded connection = 4.88cm 2 Length of longest bottom chord member = 1200mm Therefore, all bottom chord members should be able to resist an axial tensile load of 22.189 kN and a possible reversal of stresses with a compressive load of 12.896 kN Therefore ultimate design force in the member = Fu = γ GjGk + γ WkWk = G k + 1.5W k. Partial factor for leading variable actions (W k) = γW k = 1.5 Partial factor for permanent actions (DK) = γGj = 1.0 (favourable) LOAD CASE 2: DEAD LOAD + WIND LOAD acting simultaneously Ultimate design force (N Ed) = 1.35G k + 1.5Q k LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only In all cases, (T) – Tensile force (C) – Compressive force Structural Design of Roof Trusses to Eurocode 3Īll structural steel employed in the design has the following properties į u (ultimate tensile strength = 430 N/mm 2)ĭesign of the bottom chord (considering maximum effects) F 2-3 = F 3-2 so kindly distinguish this from other numeric elements N/B: Please note that the internal forces in the members are denoted by F i-j which is also equal to F j-i e.g. To see how wind load is analysed using Eurocode, click HEREĪnalysis of the Truss for Internal Forces Therefore the nodal wind load (W k) = 1.08 kN/m 2 × 1.2m × 3m = 3.888 kN Vertical component p ev = q e cos θ = 1.35 × cos 36.869 = 1.08 kN/m 2 acting upwards ↑ Therefore the external wind pressure normal to the roof is When the wind is blowing from right to left, the resultant pressure coefficient on the windward and leeward slopes with positive internal pressure ( c pe) is taken as −0.9 Wind velocity pressure (dynamic) is assumed as = q p(z) = 1.5 kN/m 2 Therefore the nodal variable load (Q K) = 0.75 kN/m 2 × 1.2m × 3m = 2.7 kN Therefore the nodal permanent load (g k) = 0.536 kN/m 2 × 1.2m × 3m = 1.9296 kNĬategory of roof = Category H – Roof not accessible except for normal maintenance and repairs (Table 6.9 EN :2001) Self weight of trusses (assume) = 0.2 kN/m 2 Weight of ceiling (adopt 10mm insulation fibre board) = 0.077 kN/m 2 Self-weight of long span aluminium roofing sheet (0.55mm gauge thickness) = 0.019 kN/m 2 It is desired to specify the appropriate angle sections that will safely carry the anticipated loading using Eurocode design code (Specified steel grade S 275). The truss is made up of Howe Truss configuration spaced at 3m intervals. The skeletal structure of a roof system (18.0m long and 7.2m wide) is as shown in the Figure below. To illustrate this, a simple design example has been presented. Example on the Design of Steel Roof Trusses The flexural buckling in Eurocode 3 is achieved by applying a reduction factor to the compression resistance. In most truss members, only flexural buckling of the compressed members in the plane of the truss structure and out of the plane of the truss structure need to be evaluated. However, it should be noted that the architectural design of the building determines its external geometry and governs the slope given to the top chord of the truss.įor the design of a compression member in a roof truss, several buckling modes need to be considered. For a good structural performance of roof trusses, the ratio of span to truss depth should be in the range of 10 to 15.
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